on the closed interval [0, 2pi], find the max value of the function f(x) = 4 sin x - 3 cos x

Question

a. 3

b. 4

c. 24/5

d. 5

e. none of these

I went about solving this by:

1. finding f'(x): I got 4 cos (x) + 3 sin (x)
2. then setting f'(x) to zero: 0 = 4 cos (x) + 3 sin (x)
3. simplifying the equation to: sin(x)/cos(x) = -4/3
4. further simplifying I got: tan(x) = -4/3
5. I got x could = about 127 and 306 degrees, I tried setting up a sign chart but the interval is always deceasing
6. I'm not sure how to solve this?

Answer 1

Let α and α + π be the second/forth quadrant angles (respectively) such that tanα = -4/3.

The first derivative is positive in the intervals x ∈ [0, α] and [α, α+ π], so x = α is a point of relative maximum.

To determine the absolute value of f in the [0, 2pi] interval, we need to compare f(α) with the values f takes in the interval limits, x = 0 and x = 2pi

f(0) = -3 = f(2\pi) since the function is 2pi periodic

f(α) = cos(α) * (4 tanα -3)

Now, remember that tanα = sinα/cosα and that sin^2α + cos^2α = 1, and you can find cos(α) = -\sqrt{1/(tan^2α + 1)}. Now all we need is tanα = -4/3:

f(α) = 5

So, 5 is the maximal value of that function in the given interval.

Answer 2

the second quadrant angle will result in

function value of 4(4/5)-3(-3/5)=5

if tan(x)=4/(-3), then sin(x)=4/5 and cos(x)= -3/5

5 is the hypotenuse of the right triangle with

y value of 4 and x value of -3